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Date	May 2019	Marks available	2	Reference code	19M.1.SL.TZ1.S_8
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Show that	Question number	S_8	Adapted from	N/A

Question

Let $f(x) = 9 - {x^2}$ , $x \in \mathbb{R}$ .

The following diagram shows part of the graph of $f$ .

Rectangle PQRS is drawn with P and Q on the $x$ -axis and R and S on the graph of $f$ .

Let OP = $b$ .

Consider another function $g(x) = {\left( {x - 3} \right)^2} + k$ , $x \in \mathbb{R}$ .

Find the $x$ -intercepts of the graph of $f$ .

[2]

Show that the area of PQRS is $18b - 2{b^3}$ .

[2]

Hence find the value of $b$ such that the area of PQRS is a maximum.

[5]

Show that when the graphs of $f$ and $g$ intersect, $2{x^2} - 6x + k = 0$ .

[2]

Given that the graphs of $f$ and $g$ intersect only once, find the value of $k$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $f(x) = 0$ , $9 - {x^2} = 0$ , one correct solution

$x = - 3$ , 3 (accept (3, 0), (−3, 0)) A1 N2

[2 marks]

valid approach (M1)

eg height = $f(b)$ , base = 2(OP) or $2b$ , $2b\left( {9 - {x^2}} \right)$ , $2b \times f(b)$

correct working that clearly leads to given answer A1

eg $2b\left( {9 - {b^2}} \right)$

Note: Do not accept sloppy notation eg $2b \times 9 - {b^2}$ .

area = $18b - 2{b^3}$ AG N0

[2 marks]

setting derivative = 0 (seen anywhere) (M1)

eg $A' = 0$ , ${\left[ {18b - 2{b^3}} \right]^\prime } = 0$

correct derivative (must be in terms of $b$ only) (seen anywhere) A2

eg $18 - 6{b^2}$ , $2b\left( { - 2b} \right) + \left( {9 - {b^2}} \right) \times 2$

correct working (A1)

eg $6{b^2} = 18$ , $b = \pm \sqrt 3$

$b = \sqrt 3$ A1 N3

[5 marks]

valid approach (M1)

eg $f = g$ , $9 - {x^2} = {\left( {x - 3} \right)^2} + k$

correct working (A1)

eg $9 - {x^2} = {x^2} - 6x + 9 + k$ , $9 - {x^2} - {x^2} + 6x - 9 - k = 0$

$2{x^2} - 6x + k = 0$ AG N0

[2 marks]

METHOD 1 (discriminant)

recognizing to use discriminant (seen anywhere) (M1)

eg Δ, ${b^2} - 4ac$

discriminant = 0 (seen anywhere) M1

correct substitution into discriminant (do not accept only in quadratic formula) (A1)

eg ${\left( { - 6} \right)^2} - 4\left( 2 \right)\left( k \right)$ , ${\left( 6 \right)^2} - 4\left( 2 \right)\left( k \right)$

correct working (A1)

eg $36 - 8k = 0$ , $8k = 36$

$k = \frac{{36}}{8}\,\,\,\,\left( { = \frac{9}{2}{\text{,}}\,\,4.5} \right)$ A1 N2

METHOD 2 (completing the square)

valid approach to complete the square (M1)

eg $2\left( {{x^2} - 3x + \frac{9}{4}} \right) = - k + \frac{{18}}{4}$ , ${x^2} - 3x + \frac{9}{4} - \frac{9}{4} + \frac{k}{2} = 0$

correct working (A1)

eg $2{\left( {x - \frac{3}{2}} \right)^2} = - k + \frac{{18}}{4}$ , ${\left( {x - \frac{3}{2}} \right)^2} - \frac{9}{4} + \frac{k}{2} = 0$

recognizing condition for one solution M1

eg ${\left( {x - \frac{3}{2}} \right)^2} = 0$ , $- \frac{9}{4} + \frac{k}{2} = 0$

correct working (A1)

eg $- k = \frac{{18}}{4}$ , $\frac{k}{2} = \frac{9}{4}$

$k = \frac{{18}}{4}\,\,\,\,\left( { = \frac{9}{2}{\text{,}}\,\,4.5} \right)$ A1 N2

METHOD 3 (using vertex)

valid approach to find vertex (seen anywhere) M1

eg ${\left( {2{x^2} - 6x + k} \right)^\prime } = 0$ , $- \frac{b}{{2a}}$

correct working (A1)

eg ${\left( {2{x^2} - 6x + k} \right)^\prime } = 4x - 6$ , $- \frac{{\left( { - 6} \right)}}{{2\left( 2 \right)}}$

$x = \frac{6}{4}\,\,\left( { = \frac{3}{2}} \right)$ (A1)

correct substitution (A1)

eg $2{\left( {\frac{3}{2}} \right)^2} - 6\left( {\frac{3}{2}} \right) + k = 0$

$k = \frac{{18}}{4}\,\,\,\,\left( { = \frac{9}{2}{\text{,}}\,\,4.5} \right)$ A1 N2

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » SL 5.7—Optimisation

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EXN.2.SL.TZ0.2e:
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Solve $\frac{d L}{d x} = 0$
21N.1.SL.TZ0.12a:
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21N.1.SL.TZ0.12b.ii:
Interpret your answer to (b)(i) in context.

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